Correct record of fractional number does not contain irrationality in a denominator. Such record is also easier perceived by sight therefore at emergence of irrationality in a denominator it is reasonable to get rid of it. In this case the irrationality can pass into numerator.
Instruction
1. For a start it is possible to review the simplest example - 1/sqrt (2). A square root from two - irrational number in a denominator. In this case it is necessary to multiply numerator and a denominator of fraction on its denominator. It will provide a rational number in a denominator. Really, sqrt(2)*sqrt(2) = sqrt(4) = 2. Multiplication of two identical square roots at each other will give as a result what is under each of roots: in this case - the two. As a result: 1/sqrt (2) = (1*sqrt (2)) / (sqrt (2) *sqrt (2)) = sqrt(2)/2. This algorithm approaches also fractions in which denominator the root is multiplied by a rational number. The numerator and a denominator in this case need to be increased by the root which is in a denominator. Example: 1/(2*sqrt (3)) = (1*sqrt (3)) / (2*sqrt (3) *sqrt(3)) = sqrt(3)/(2*3) = sqrt(3)/6.
2. It is absolutely similarly necessary to work if in a denominator there is not a square root, and, we will tell cubic or any other degree. The root in a denominator needs to be multiplied by just the same root, to multiply by the same root also numerator. Then the root will pass into numerator.
3. At more difficult case in a denominator there is a sum or a rational number or two irrational numbers. In case of the sum (difference) of two square roots or a square root and a rational number it is possible to use a well-known formula (x+y) (x-y) = (x^2) (-y^2). It will help to get rid of irrationality in a denominator. If in a denominator a difference, it is necessary to multiply numerator and a denominator for the sum of the same numbers if the sum - on a difference. This multiplied sum or a difference will be called interfaced to the expression standing in a denominator. The effect is well visible to this scheme on an example: 1/(sqrt(2)+1) = (sqrt(2)-1)/(sqrt(2)+1) (sqrt(2)-1) = (sqrt(2)-1) / ((sqrt(2) ^2) - (1^2)) = (sqrt(2)-1)/(2-1) = sqrt(2)-1.
4. If at a denominator there is a sum (difference) at which there is a root of bigger degree, then the situation becomes uncommon and disposal of irrationality in a denominator is not always possible